Block of Ice, Copper, change of temperatures and calorimeter. Can you please help me w/this problem? Thank You

Question

A 40 g block of ice is cooled to -68°C. and is then added to 620 g of water in an 80 g copper calorimeter at a temperature of 22°C. Determine the final temperature of the system consisting of the ice, water, and calorimeter. Remember that the ice must first warm to 0°C, melt, and then continue warming as water. The specific heat of ice is 0.500 cal/g °C = 2090 J/kg°C.
_________°C

Answer 1

The eqtn (w cp del T) cold side =
Sum ( w cp del T)high side + Phase change effects.
ALL NUMBERS APPROXIMATE
The copper calorimeter has a very low cp, and can be ignored. To get everything to zero C, we need to supply 40 x 2.09 J/gmC x 68 degC about 5500 J to the ice and we can remove 52000 J from the water. This is about 2400 J/degree heat supply from the calorimeter water. To melt the ice, we need 334J/g. To melt 40 g, we need 13200 J, which is still less than can be supplied by cooling water.

To figure a final temperature, we need to recast our problem. We know that we need 19000 J to melt the super-cold ice. From our analysis, can be supplied by cooling the calorimeter by about 8 degrees (19000/2400). So we now heat 40 g of zero degree water with 620 g of 14 deg water, and seek the final temperature , which will be around 13 deg C.